Question: $g(t) = 4t$ $f(n) = -3n^{2}-7n-3(g(n))$ $ g(f(-1)) = {?} $
Explanation: First, let's solve for the value of the inner function, $f(-1)$ . Then we'll know what to plug into the outer function. $f(-1) = -3(-1)^{2}+(-7)(-1)-3(g(-1))$ To solve for the value of $f$ , we need to solve for the value of $g(-1)$ $g(-1) = (4)(-1)$ $g(-1) = -4$ That means $f(-1) = -3(-1)^{2}+(-7)(-1)+(-3)(-4)$ $f(-1) = 16$ Now we know that $f(-1) = 16$ . Let's solve for $g(f(-1))$ , which is $g(16)$ $g(16) = (4)(16)$ $g(16) = 64$